Trees on the level

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 591 Accepted Submission(s): 200

Problem Description

Trees are fundamental in many branches of computer science. Current state-of-the art parallel computers such as Thinking Machines' CM-5 are based on fat trees. Quad- and octal-trees are fundamental to many algorithms in computer graphics.

This problem involves building and traversing binary trees.
Given a sequence of binary trees, you are to write a program that prints a level-order traversal of each tree. In this problem each node of a binary tree contains a positive integer and all binary trees have have fewer than 256 nodes.

In a level-order traversal of a tree, the data in all nodes at a given level are printed in left-to-right order and all nodes at level k are printed before all nodes at level k+1.

For example, a level order traversal of the tree

is: 5, 4, 8, 11, 13, 4, 7, 2, 1.

In this problem a binary tree is specified by a sequence of pairs (n,s) where n is the value at the node whose path from the root is given by the string s. A path is given be a sequence of L's and R's where L indicates a left branch and R indicates a right branch. In the tree diagrammed above, the node containing 13 is specified by (13,RL), and the node containing 2 is specified by (2,LLR). The root node is specified by (5,) where the empty string indicates the path from the root to itself. A binary tree is considered to be completely specified if every node on all root-to-node paths in the tree is given a value exactly once.

Input

The input is a sequence of binary trees specified as described above. Each tree in a sequence consists of several pairs (n,s) as described above separated by whitespace. The last entry in each tree is (). No whitespace appears between left and right parentheses.

All nodes contain a positive integer. Every tree in the input will consist of at least one node and no more than 256 nodes. Input is terminated by end-of-file.

Output

For each completely specified binary tree in the input file, the level order traversal of that tree should be printed. If a tree is not completely specified, i.e., some node in the tree is NOT given a value or a node is given a value more than once, then the string ``not complete'' should be printed

Sample Input

(11,LL) (7,LLL) (8,R) (5,) (4,L) (13,RL) (2,LLR) (1,RRR) (4,RR) () (3,L) (4,R) ()

Sample Output

5 4 8 11 13 4 7 2 1

not complete

题意：给出给出二叉树上的点和它的位置，但有可能不能构成二叉树，例如没根节点或者同一位置有多个点存在。

思路：先建树，判断其能否构成二叉树，若能，则用bfs层次遍历该二叉树

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;

const int maxn = 256 + 10;

struct Node{
  bool have_value;
  int v;
  Node* left, *right;
  Node():have_value(false),left(NULL),right(NULL){}
};

Node* root;

Node* newnode() { return new Node(); }

bool failed;
void addnode(int v, char* s) {
  int n = strlen(s);
  Node* u = root;
  for(int i = 0; i < n; i++)
    if(s[i] == 'L') {
      if(u->left == NULL) u->left = newnode();
      u = u->left;
    } else if(s[i] == 'R') {
      if(u->right == NULL) u->right = newnode();
      u = u->right;
    }
  if(u->have_value) failed = true;
  u->v = v;
  u->have_value = true;
}

void remove_tree(Node* u) {
  if(u == NULL) return;
  remove_tree(u->left);
  remove_tree(u->right);
  delete u;
}

char s[maxn];
bool read_input() {
  failed = false;
  remove_tree(root);
  root = newnode();
  for(;;) {
    if(scanf("%s", s) != 1) return false;
    if(!strcmp(s, "()")) break;
    int v;
    sscanf(&s[1], "%d", &v);
    addnode(v, strchr(s, ',')+1);
  }
  return true;
}

bool bfs(vector<int>& ans) {
  queue<Node*> q;
  ans.clear();
  q.push(root);
  while(!q.empty()) {
    Node* u = q.front(); q.pop();
    if(!u->have_value) return false;
    ans.push_back(u->v);
    if(u->left != NULL) q.push(u->left);
    if(u->right != NULL) q.push(u->right);
  }
  return true;
}

int main() {
  vector<int> ans;
  while(read_input()) {
    if(!bfs(ans)) failed = 1;
    if(failed) printf("not complete
");
    else {
      for(int i = 0; i < ans.size(); i++) {
        if(i != 0) printf(" ");
        printf("%d", ans[i]);
      }
      printf("
");
    }
  }
  return 0;
}

2018-04-10